📂Abstract Algebra

Rules for Multiplication in a Ring

Theorem

Given that $a,\ b,\ c$ is an element of the ring $R$ and $0$ is the identity element for addition, the following properties hold:

1. $a0=0a=0$
2. $a(-b)=(-a)b=-(ab)$
3. $(-a)(-b)=ab$
4. $a(b-c)=ac-ac \ \ \And\ \ (b-c)a=ba-ca$ If there exists a multiplicative identity element $1$, then the following properties also hold:
5. $(-1)a=-a$
6. $(-1)(-1)=1$

Proof

1.

It’s about the property that multiplying any element with the additive identity still results in the additive identity. $$a0=a(0+0)=a0+a0$$ Since $a0$ is an element of the ring $R$, it has an additive inverse. Hence, by adding $-a0$ to both sides according to the Cancellation law, $$a0-a0=a0+a0-a0 \implies 0=a0$$ Similarly, $$0a=(0+0)a=0a+0a \implies 0a-0a=0a+0a-0a \implies 0=0a$$ Therefore, it is $a0=0a=0$.

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2.

Since the ring $R$ follows the distributive law for multiplication, $$ab+a(-b)=a\left( b + (-b) \right)=a0$$ According to Theorem 1, since it’s $a0=0$, $$ab+a(-b)=0$$ Adding $a(-b)$ to $ab$ yields the identity element $0$, so $a(-b)$ is the additive identity for $ab$. That is, $$a(-b)=-(ab)$$ And similarly, $$ab+(-a)b=(a-a)b=0b=0$$ leads to $(-a)b=-(ab)$. Finally, the following is obtained. $$a(-b)=(-a)b=-(ab)=-ab$$

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3.

By Theorem 2, $$(-a)(-b) =-\left( (-a)b\right) =-\left( -(ab) \right)$$ Since the inverse of an inverse is itself, the following is obtained. $$(-a)(-b)=ab$$

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4.

$$a(b-c)=a(b+(-c))=ab+a(-c)$$ Since it’s $a(-c)=-(ac)=-ac$ by Theorem 2, $$a(b-c)=ab-ac$$ is obtained. Similarly, the following is obtained. $$(b-c)a=ba+(-c)a=ba-ca$$

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5.

Since it’s $(-1)a=-(1a)$ by Theorem 2, and $1$ is the multiplicative identity, it is $1a=a$, and therefore, it’s $(-1)a=-a$.

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6.

As in the proof of Theorem 3, by repeatedly applying Theorem 2, the following is obtained. $$(-1)(-1)=-(1(-1))=-(-(1\cdot 1))=-(-1)=1$$

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